Problem: Multiply the following complex numbers: $({4-3i}) \cdot ({-1-i})$
Answer: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({4-3i}) \cdot ({-1-i}) = $ $ ({4} \cdot {-1}) + ({4} \cdot {-1}i) + ({-3}i \cdot {-1}) + ({-3}i \cdot {-1}i) $ Then simplify the terms: $ (-4) + (-4i) + (3i) + (3 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ -4 + (-4 + 3)i + 3i^2 $ After we plug in $i^2 = -1$ , the result becomes $ -4 + (-4 + 3)i - 3 $ The result is simplified: $ (-4 - 3) + (-1i) = -7-i $